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True [87]
4 years ago
10

Factor. 25m100−121n16 (5m50−11n8)(5m50+11n8) (5m50−11n8)2 (5m10−11n4)2 (5m10−11n4)(5m10+11n4)

Mathematics
1 answer:
Flura [38]4 years ago
3 0

Answer:

The answer to your question is (5m50−11n8)(5m50+11n8)

Step-by-step explanation:

                                             25m¹⁰⁰−121n¹⁶

find the prime factors of 25 and 100

    25     5                                  121     11

       5     5                                   11      11

        1                                            1

25 = 5 x 5 = 5²                       121 = 11 x 11 = 11²

m¹⁰⁰ = (m⁵⁰)²                           n¹⁶ = (n⁸)²

Substitution

                     25m¹⁰⁰−121n¹⁶ = 5²(m⁵⁰)²   - 11² (n⁸)²

                                            = (5m⁵⁰  - 11n⁸)  (5m⁵⁰  + 11n⁸)

(5m50−11n8)(5m50+11n8)

(5m50−11n8)2

(5m10−11n4)2

(5m10−11n4)(5m10+11n4)

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Answer: Choice B. The vertex is (6,-4)

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Work Shown:

Step 1 is to expand out (x-8)(x-4) using the FOIL rule or the box method or the distribution rule

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(x-8)(x-4) = x*x+x*(-4)-8*x-8*(-4)

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So (x-8)(x-4) turns into x^2-12x+32

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-----------------

Use the values of a & b to find the value of h, which is the x coordinate of the vertex

h = -b/(2*a)

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h = 6

Then this is plugged back into the original function to find the y coordinate of the vertex. We can use either (x-8)(x-4) or x^2-12x+32 since they are equivalent expressions

k = y coordinate of vertex

k = f(h) = f(6) since h = 6

f(x) = (x-8)(x-4)

f(6) = (6-8)(6-4)

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note that

f(x) = x^2-12x+32

f(6) = (6)^2-12(6)+32

f(6) = 36-72+32

f(6) = -36+32

f(6) = -4

So we get the same result using either expression

So k = f(h) = f(6) = -4

Since h = 6 and k = -4, the vertex is (h,k) = (6,-4). So that's why the answer is choice B.

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